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Home Partial Differentiation Maxima and Minima Examples Example 1: Closed Rectangle
 
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Example 1: Closed Rectangle

Find the maximum and minimum of

z = x2 + y2 - xy - x

on the closed rectangle

0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

Step 1

The region D is sketched in Figure 11.7.4.

11_partial_differentiation-459.gif

Figure 11.7.4

Step 2

11_partial_differentiation-457.gif = 2x - y - 1, 11_partial_differentiation-458.gif = 2y - x.

Step 3

2x - y - 1 = 0, 2y - x = 0.

Solving for x and y we get one critical point

y = ⅓, x = ⅔

The value of z at that point is

z = (⅔)2 + (⅓)2 - ⅔ · ⅓ - ⅔ = -⅓

Step 4

We make a table.

Boundary Line

z

Maximum

Minimum

x = 0, 0 ≤ y ≤ 1

y2

1 at(0, 1)

0 at (0,0)

x =1, 0 ≤ y ≤ 1

y2 - y

0 at corners

-¼ at (1, ½)

y = 0, 0 ≤ x ≤ 1

x2 - x

0 at corners

-¼ at (½, 0)

y =1, 0 ≤ x ≤ 1

x2 + 1 - 2x

1 at (0,1)

0 at (1, 1)

 

The values from Steps 3 and 4 are also shown on the sketch of D in Figure 11.7.5.

11_partial_differentiation-460.gif

Figure 11.7.5

CONCLUSION

Maximum: z = 1 at (0, 1) Minimum: z = -⅓ at(⅔, ⅓).

The maximum is at a boundary point and the minimum at an interior point.
Home Partial Differentiation Maxima and Minima Examples Example 1: Closed Rectangle

Last Update: 2006-11-15